I have an exercise that asks to compute the triple integral over a region $E$ (a paraboloid with a horizontal slice), which I know to be
$$\iiint_E x\,dV=\int_{-1}^1\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\int_{6y^2+6z^2}^6x\,dx\,dz\,dy=12\pi$$
I computed the result via a conversion to cylindrical coordinates at first by using the following parameters:
$$\begin{cases}x=x\\
y=r\sin t\\
z=r\cos t\end{cases}$$
To compute the determinant of the Jacobian, I have
$$|J|=\begin{vmatrix}
\frac{\partial x}{\partial x}&\frac{\partial y}{\partial x}&\frac{\partial z}{\partial x}\\
\frac{\partial x}{\partial r}&\frac{\partial y}{\partial r}&\frac{\partial z}{\partial r}\\
\frac{\partial x}{\partial t}&\frac{\partial y}{\partial t}&\frac{\partial z}{\partial t}
\end{vmatrix}=
\begin{vmatrix}1&0&0\\0&\sin t&\cos t\\0&r\cos t&-r\sin t\end{vmatrix}=-r$$
However, if I were to swap $x$ and $z$ (so that the paraboloid were oriented along the $z$ axis in place of the $x$ axis) and use the usual parameters,
$$\begin{cases}x=r\cos t\\y=r\sin t\\z=z\end{cases}$$
I get the usual Jacobian, $|J|=r$. I felt I trusted this result more, so I stuck with this one to get $12\pi$.
Why is there a discrepancy here? Does the order of the partial derivatives in the Jacobian matter?
Best Answer
Indeed you must take the absolute value of the jacobian. Both the change of variables are correct.
Take a closer look to the different diffeomorphisms of the two changes of variables. They are $$\begin{array}{ccc} \Phi:& \mathbb R^3& \to & \mathbb R^3\\ & \begin{pmatrix} r\\t\\z\end{pmatrix}&\mapsto & \begin{pmatrix} r\cos(t)\\r\sin(t)\\z\end{pmatrix}\end{array} $$ $$\begin{array}{ccc} \Psi:& \mathbb R^3& \to & \mathbb R^3\\ & \begin{pmatrix} r\\t\\x\end{pmatrix}&\mapsto & \begin{pmatrix} x\\r\sin(t)\\r\cos(t)\end{pmatrix}\end{array} $$
As you can see, $$A\Phi((r,t,z)^T)= \begin{pmatrix}0&0&1\\0&1&0\\1&0&0 \end{pmatrix} \begin{pmatrix} r\cos(t) \\r\sin(t) \\z \end{pmatrix} = \begin{pmatrix} z\\r\sin(t)\\r\cos(t)\end{pmatrix} = \Psi((r,t,z)^T) $$
The matrix $A$ has determinant $-1$ and is a simmetry of $\mathbb R^3$ in itself. Since $A$ does not depend on the variables and because of Binet's theorem, $$J_{\Psi}=AJ_{\Phi}\implies \text{det}(J_{\Psi})=-\text{det}(J_{\Phi})$$ Since we usually want volume to be positive we usually take $|\text{det}(J)|$ over $\text{det}(J)$ as you did: this follows intuitively if you believe that symmetrical shapes have the same volume.
Moreover, a simmetry flips the normal vectors of surfaces in certain directions. For instance ,the normal vector $(0,1)$ on top of a circle in $\mathbb R^2$ becomes $(0,-1)$ when you apply the simmetry $(x,y)\mapsto(x,-y)$.
A minus sign in your calculations of $\text {det}(J)$ shows that you have inverted the orientation of your surface while changing coordinates, so you have "flipped" some normal vectors. Nothing to worry about when calculating volumes!